By these calculations temp drop might explain "Deflategate"

[Blitzer88]

So much sciencey stuff..........................my brain hurts...................

 

[bmorepunk]

So much sciencey stuff..........................my brain hurts...................

It's basic addition, subtraction, multiplication and division.

We're not even doing basic calculus here, kids.

 

[mikeak]

They did check Indy's balls, and they were correct pressure

Just posted this in the other thread but you don't know what psi the Colts balls were inflated to prior. If they were at the upper limit then they would be measure as legal despite some deflation

 

[Vancanhawksfan]

Maybe somebody smarter than me can check these calculations, but it seems to me that there is a possibility that the ball pressure discrepancy could be explained if the referees originally tested the balls @ 12.5 psi in a heated locker room of 69.57 F, but were re-checked at half-time at 11 psi at a recorded outside temperature of 52 F...in other words... if the referees F'd up.

Here are my calculations...

Assumptions:

1. Let's assume that all balls tested were done pregame in the referees locker room just a little before 2hrs before gametime (4:53pm or 16:53) at 12.5 psi;
(NFL rules state that balls are turned over to teams 2 hrs 15 mins before gametime)

2. And then let's also assume all "deflated" balls were tested outside and found to be 11 psi at 7:53 pm EST (19:53)

Weather Observations January 18, 2015 - Foxboro, Massachusetts

http://www.weatherforyou.com/reports/in ... &icao=KOWD

@ 16:53
Temp in referees locker room: to be determined
Atmospheric air pressure: 29.83 in

@ 19:53
Outside temperature: 52F = 284.61K
Atmospheric air pressure: 29.68 in

Ideal Gas Law

PV = nRT

rewritten as P/T = nR/V = constant

Therefore:
P(1)/T(1) = P(2)/T(2)

(29.83 in + 11 in)/284.61K = (29.68 in + 12.5 in)/T(2)

T(2) = 295.10K = 69.57 F = possible referee locker room temperature

It seems to me that it is very realistic that the ref's locker room temperature was 70 F. Let's face it - refs aren't physics majors and they may very well not have considered that the balls would have had to be warmed to the ref's locker room temperature before being retested again.

Adding barometric pressure to the calculation certainly takes into effect that the kind of pressure gauge used to check the balls are relative. However, you have made a huge mistake in adding barometric pressure (which is measured in inHg) and psi together. You have to convert inHg to psi first.

29.83 inHg = 14.65 psi
29.68 inHg = 14.58 psi

Difference in measurement due to barometric pressure: 14.65 psi - 14.58 psi = 0.07 psi (and I'm spotting you the round up here). The barometric pressure's effect is two orders of magnitude below that of a 2 psi measurement, and really not worth putting into the calculation for something like this.

The temperature drop to account for 2 psi is:

P1/T1 = P2/T2
T2 = (T1*P2) / P1 = (295.37 Kelvin * 72,394.95 pascals) /  86,184.47 pascals = 248.11 Kelvin

which is about -13 F.

The resulting pressure, if started at 12.5 psi, going from 72 F (295.37 K) to 52 F (284.26) would be:

P2 = (P1*T2) / T1
P2 = (86,184.47 pascals * 284.26 K) / 295.37 K
P2 = 82942.74 pascals

That's about 12.0 psi.

Thank you for addressing the question at hand. It is much more appreciated than the name calling I'm getting from some of the others here.

So I guess to correct myself then, to go from 12.5 psi to 11 psi at 52 F, the room temperature of the ref's room would need to be around 81 degrees? I suppose that's not plausible unless there was something else at play?

 

[mikeak]

^ they were in the hot tub when they checked the footballs

 

[bmorepunk]

Maybe somebody smarter than me can check these calculations, but it seems to me that there is a possibility that the ball pressure discrepancy could be explained if the referees originally tested the balls @ 12.5 psi in a heated locker room of 69.57 F, but were re-checked at half-time at 11 psi at a recorded outside temperature of 52 F...in other words... if the referees F'd up.

Here are my calculations...

Assumptions:

1. Let's assume that all balls tested were done pregame in the referees locker room just a little before 2hrs before gametime (4:53pm or 16:53) at 12.5 psi;
(NFL rules state that balls are turned over to teams 2 hrs 15 mins before gametime)

2. And then let's also assume all "deflated" balls were tested outside and found to be 11 psi at 7:53 pm EST (19:53)

Weather Observations January 18, 2015 - Foxboro, Massachusetts

http://www.weatherforyou.com/reports/in ... &icao=KOWD

@ 16:53
Temp in referees locker room: to be determined
Atmospheric air pressure: 29.83 in

@ 19:53
Outside temperature: 52F = 284.61K
Atmospheric air pressure: 29.68 in

Ideal Gas Law

PV = nRT

rewritten as P/T = nR/V = constant

Therefore:
P(1)/T(1) = P(2)/T(2)

(29.83 in + 11 in)/284.61K = (29.68 in + 12.5 in)/T(2)

T(2) = 295.10K = 69.57 F = possible referee locker room temperature

It seems to me that it is very realistic that the ref's locker room temperature was 70 F. Let's face it - refs aren't physics majors and they may very well not have considered that the balls would have had to be warmed to the ref's locker room temperature before being retested again.

Adding barometric pressure to the calculation certainly takes into effect that the kind of pressure gauge used to check the balls are relative. However, you have made a huge mistake in adding barometric pressure (which is measured in inHg) and psi together. You have to convert inHg to psi first.

29.83 inHg = 14.65 psi
29.68 inHg = 14.58 psi

Difference in measurement due to barometric pressure: 14.65 psi - 14.58 psi = 0.07 psi (and I'm spotting you the round up here). The barometric pressure's effect is two orders of magnitude below that of a 2 psi measurement, and really not worth putting into the calculation for something like this.

The temperature drop to account for 2 psi is:

P1/T1 = P2/T2
T2 = (T1*P2) / P1 = (295.37 Kelvin * 72,394.95 pascals) /  86,184.47 pascals = 248.11 Kelvin

which is about -13 F.

The resulting pressure, if started at 12.5 psi, going from 72 F (295.37 K) to 52 F (284.26) would be:

P2 = (P1*T2) / T1
P2 = (86,184.47 pascals * 284.26 K) / 295.37 K
P2 = 82942.74 pascals

That's about 12.0 psi.

Thank you for addressing the question at hand. It is much more appreciated than the name calling I'm getting from some of the others here.

So I guess to correct myself then, to go from 12.5 psi to 11 psi at 52 F, the room temperature of the ref's room would need to be around 81 degrees? I suppose that's not plausible unless there was something else at play?

The temperature change alone is not enough to explain a supposed 2 psi difference. The other things that have come up about this seem improbable, but I'm not in the profession of handling footballs.

 

[Vancanhawksfan]

Maybe somebody smarter than me can check these calculations, but it seems to me that there is a possibility that the ball pressure discrepancy could be explained if the referees originally tested the balls @ 12.5 psi in a heated locker room of 69.57 F, but were re-checked at half-time at 11 psi at a recorded outside temperature of 52 F...in other words... if the referees F'd up.

Here are my calculations...

Assumptions:

1. Let's assume that all balls tested were done pregame in the referees locker room just a little before 2hrs before gametime (4:53pm or 16:53) at 12.5 psi;
(NFL rules state that balls are turned over to teams 2 hrs 15 mins before gametime)

2. And then let's also assume all "deflated" balls were tested outside and found to be 11 psi at 7:53 pm EST (19:53)

Weather Observations January 18, 2015 - Foxboro, Massachusetts

http://www.weatherforyou.com/reports/in ... &icao=KOWD

@ 16:53
Temp in referees locker room: to be determined
Atmospheric air pressure: 29.83 in

@ 19:53
Outside temperature: 52F = 284.61K
Atmospheric air pressure: 29.68 in

Ideal Gas Law

PV = nRT

rewritten as P/T = nR/V = constant

Therefore:
P(1)/T(1) = P(2)/T(2)

(29.83 in + 11 in)/284.61K = (29.68 in + 12.5 in)/T(2)

T(2) = 295.10K = 69.57 F = possible referee locker room temperature

It seems to me that it is very realistic that the ref's locker room temperature was 70 F. Let's face it - refs aren't physics majors and they may very well not have considered that the balls would have had to be warmed to the ref's locker room temperature before being retested again.

Adding barometric pressure to the calculation certainly takes into effect that the kind of pressure gauge used to check the balls are relative. However, you have made a huge mistake in adding barometric pressure (which is measured in inHg) and psi together. You have to convert inHg to psi first.

29.83 inHg = 14.65 psi
29.68 inHg = 14.58 psi

Difference in measurement due to barometric pressure: 14.65 psi - 14.58 psi = 0.07 psi (and I'm spotting you the round up here). The barometric pressure's effect is two orders of magnitude below that of a 2 psi measurement, and really not worth putting into the calculation for something like this.

The temperature drop to account for 2 psi is:

P1/T1 = P2/T2
T2 = (T1*P2) / P1 = (295.37 Kelvin * 72,394.95 pascals) /  86,184.47 pascals = 248.11 Kelvin

which is about -13 F.

The resulting pressure, if started at 12.5 psi, going from 72 F (295.37 K) to 52 F (284.26) would be:

P2 = (P1*T2) / T1
P2 = (86,184.47 pascals * 284.26 K) / 295.37 K
P2 = 82942.74 pascals

That's about 12.0 psi.

Thank you for addressing the question at hand. It is much more appreciated than the name calling I'm getting from some of the others here.

So I guess to correct myself then, to go from 12.5 psi to 11 psi at 52 F, the room temperature of the ref's room would need to be around 81 degrees? I suppose that's not plausible unless there was something else at play?

The temperature change alone is not enough to explain a supposed 2 psi difference. The other things that have come up about this seem improbable, but I'm not in the profession of handling footballs.

Fair enough...me neither. (also, to be fair, the change that needs to be accounted for is 1.5 psi difference). OK...I'll go back to hating on Brady and Belicek until I learn something otherwise! lol!

 

[randomation]

It was 2 psi btw not sure why you keep saying 1.5.

 

[Vancanhawksfan]

It was 2 psi btw not sure why you keep saying 1.5.

I understood that the balls were measured at 11 psi at halftime.

Regulation balls are to be between 12.5 - 13.5 psi. Brady has stated that he prefers them at 12.5 psi.

I have not heard an announcement stating what the refs measured them at. But if Brady likes them at the lower end of the scale then its safe to assume that they were at 12.5 psi when the refs originally measured them.

 

[KiwiHawk]

So I guess to correct myself then, to go from 12.5 psi to 11 psi at 52 F, the room temperature of the ref's room would need to be around 81 degrees? I suppose that's not plausible unless there was something else at play?

The difference between 295.37K and 248.11K is 47.26K. The ratio of units between Kelvin and Fahrenheit is same as the units between Celsius and Fahrenheit, which is 9/5. Therefore the temperature change required is 85.09 degrees F. If the game was played at 50F, the temperature in the referee's room would need to have been 135F. Definitely not plausible.

 

[Vancanhawksfan]

So I guess to correct myself then, to go from 12.5 psi to 11 psi at 52 F, the room temperature of the ref's room would need to be around 81 degrees? I suppose that's not plausible unless there was something else at play?

The difference between 295.37K and 248.11K is 47.26K. The ratio of units between Kelvin and Fahrenheit is same as the units between Celsius and Fahrenheit, which is 9/5. Therefore the temperature change required is 85.09 degrees F. If the game was played at 50F, the temperature in the referee's room would need to have been 135F. Definitely not plausible.

No. The ratio between Farenheit and Celsius is 9/5. Here is a conversion calculator for you for Farenheit to Kelvin:

https://www.google.ca/?gws_rd=ssl#q=con ... +to+kelvin

Plus, the difference is from 11 psi @ 284.61 K (52 F) to 12.5 psi @ 300.48 K (81 F) ...which I've already stated is unrealistic now that our friend has corrected my calculation.

 

[RunTheBall]

This guy sure seems pretty defensive for a "Seahawks" fan. It's crazy all these accounts that were made a day or two after the NFC/AFC Title Games that blindly defend a known cheating franchise, QB, and coach.

 

[Vancanhawksfan]

This guy sure seems pretty defensive for a "Seahawks" fan. It's crazy all these accounts that were made a day or two after the NFC/AFC Title Games that blindly defend a known cheating franchise, QB, and coach.

What does me discussing Tom Brady/Belichek/Patriots have anything to do with whether or not I'm a Seahawks fan? Stop being a knob, check my post history (even though its only a week) and then tell me if you still think I am.

The difference between you and I is that I don't "blindly" defend or criticize anybody which is something that you are doing to me right now. What...do you think that only a "true" Seahawks fan would only trash the Brady and company? How stupid is that?

Talk about hypocrisy.

 

[grizbob]

Already been calculated at ~80 degrees F for the temp change.

Your calculation is wrong because you didn't take into account the existing air pressure in Foxboro.

You know... you're right, cause there's a vacuum there. In other words Foxboro sucks. :lol:

 

[Narniaman]

Folks. . .

Good try on the calculations. . .

However, pV = nRT is the ideal gas law. . . for a rigid container.

If you are dealing with a gas in a container with non-linear elasticity. . . things become a whole lot more complicated.

The calculations might give you a rough guess. . . but the only way to actually determine what the pressure should be is to repeat the "experiment" using identical conditions.

One other thing to consider is that the air inside a wet football. . . might have a lower temperature than the ambient air. . . or it might have a higher temperature because of heating of the lining of the football from constant flexion.

 

[LickMyNuts]

This guy sure seems pretty defensive for a "Seahawks" fan. It's crazy all these accounts that were made a day or two after the NFC/AFC Title Games that blindly defend a known cheating franchise, QB, and coach.

What does me discussing Tom Brady/Belichek/Patriots have anything to do with whether or not I'm a Seahawks fan? Stop being a knob, check my post history (even though its only a week) and then tell me if you still think I am.

The difference between you and I is that I don't "blindly" defend or criticize anybody which is something that you are doing to me right now. What...do you think that only a "true" Seahawks fan would only trash the Brady and company? How stupid is that?

Talk about hypocrisy.

You probably don't post much because you don't take criticism well. Lighten up and enjoy the ride to the Super Bowl. It should be a great game regardless of the ball inflation.

With that said I don't know that we will ever get to the bottom of this but rest assured those balls will be filled for the Super Bowl.

 

[SPIRITOF12]

Love this portion of Belichick's presser.....

BELICHICK: There is this possibility. Now if those footballs were picked up by, let’s say swallows, and carried up several thousand feet they would be directly affected by the changing barometric pressure and temperatures of the upper atmosphere.

ESPN REPORTER: What? A swallow carrying a football?

BELICHICK: It could grip it by the stitching.

ESPN REPORTER: It's not a question of where he grips it! It's a simple question of weight ratios! A five ounce bird could not carry a one pound football.

BELICHICK: Well, it doesn't matter. Will you please put this whole thing to rest?

ESPN REPORTER: Listen. In order to maintain air-speed velocity, a swallow needs to beat its wings forty-three times every second, right?

BELICHICK: Please!

ESPN REPORTER: Am I right?

 

[Russ Willstrong]

It was 2 psi btw not sure why you keep saying 1.5.

I understood that the balls were measured at 11 psi at halftime.

Regulation balls are to be between 12.5 - 13.5 psi. Brady has stated that he prefers them at 12.5 psi.

I have not heard an announcement stating what the refs measured them at. But if Brady likes them at the lower end of the scale then its safe to assume that they were at 12.5 psi when the refs originally measured them.

Wrong. Refs statement was that the balls were under inflated by more than 15% of regulation. Lower limit of regulation air pressure is 12.5 psi which means they were near 10.5 psi but if you want to round up like what Belicheat is telling us then round up 12.5 to 13 too.

 

[KiwiHawk]

No. The ratio between Farenheit and Celsius is 9/5. Here is a conversion calculator for you for Farenheit to Kelvin:

https://www.google.ca/?gws_rd=ssl#q=con ... +to+kelvin

No that is a temperature converter between degrees Fahrenheit and degrees Kelvin, not between the units of measure. The two scales don't start at the same point.

Water freezes at 32F, but not at 32K. Water freezes at 273.15K

Using your calculator, let's explore some differences and see what the factor is:

1F = 255.928K
2F = 256.483K

Difference = 1F = 0.555K

5/9 = .555555555...

So 1 degree F = .555 or 5/9 K

Therefore, the conversion factor between units of F and K is 5/9, or if going the other way, 9/5.

By the way, you agree that the ratio between Fahrenheit and Celcius is 9/5, but did you not know that the units of C and K are exactly the same size, but that Celsius starts at the freezing point of water, whereas Kelvin starts at absolute zero, so 0C = 273.15K, 1C = 274.15K, 100C = 373.15K, etc?

You're Canadian, don't you use metric like the rest of the world except the US and Lybia?

 

[hawknation2015]

Love this portion of Belichick's presser.....

BELICHICK: There is this possibility. Now if those footballs were picked up by, let’s say swallows, and carried up several thousand feet they would be directly affected by the changing barometric pressure and temperatures of the upper atmosphere.

ESPN REPORTER: What? A swallow carrying a football?

BELICHICK: It could grip it by the stitching.

ESPN REPORTER: It's not a question of where he grips it! It's a simple question of weight ratios! A five ounce bird could not carry a one pound football.

BELICHICK: Well, it doesn't matter. Will you please put this whole thing to rest?

ESPN REPORTER: Listen. In order to maintain air-speed velocity, a swallow needs to beat its wings forty-three times every second, right?

BELICHICK: Please!

ESPN REPORTER: Am I right?

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